Generic irreducibility and the exceptional parameter lattice

In the basis of [F10], ${\mathrm{ad}}_W=\mathrm{diag}(0,2,-2)$ and ${\mathrm{ad}}_{E_{+}}{\mathrm{ad}}_{E_{-}}$ has diagonal entries $2,2,0$. The reversed composition has diagonal entries $2,0,2$ by the same brackets; each product with one $W$ and one $E_{\pm }$, or with two equal $E_{\pm }$, is off-diagonal. Thus the Killing matrix is $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$ with determinant $-128$, so $\mathfrak{g}$ is semisimple by [F15]. Put $\mathfrak{h}=\mathbb{C}W$. It is abelian, hence nilpotent; for $X=aW+b{E}_{+}+c{E}_{-}$ the brackets in [F10] give $[X,W]=-2b{E}_{+}+2c{E}_{-}$, so $N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$ and [F14] makes $\mathfrak{h}$ a Cartan subalgebra. Its root spaces are $\mathbb{C}{E}_{\pm }$ with roots $\pm \alpha$, where $\alpha (W)=2$; by [F14] they form the reduced root system of type $A_1$. Choose $\alpha$ positive. Since $B(W,W)=8$, the Killing-dual vector is $H_{\alpha }=W\mathrm{/}4$ and $\alpha (H_{\alpha })=1\mathrm{/}2$, so ${\alpha }^{\vee }=W$. The fundamental weight has $\omega (W)=1$ by [F16], hence $\alpha =2\omega$ and $\rho =\alpha \mathrm{/}2=\omega$. The root reflection in [F14] sends $t=\lambda (W)$ to $-t$.

Let $M$ be an algebraic $(\mathfrak{g},K)$-submodule and let $0\mathrm{\ne }v={\sum }_{r\in S}{a}_r{f}_r\in M$, where $S$ is its finite support and every $a_r\mathrm{\ne }0$. If $S$ has more than one element, put $q=1+{\sum }_{\{r,s\}\subset S,r\mathrm{\ne }s}\mathrm{\mid }r-s\mathrm{\mid }$; then $q>\mathrm{\mid }r-s\mathrm{\mid }$ for all distinct $r,s\in S$, so the eigenvalues $e^{2\pi ir\mathrm{/}q}$ of $k_{2\pi \mathrm{/}q}$ on these modes are distinct. Lagrange interpolation in $\mathrm{\Pi }(k_{2\pi \mathrm{/}q})$ extracts each $a_r{f}_r$ as a finite linear combination of $K$-translates of $v$, hence $f_r\in M$; for singleton support this is immediate. Thus every nonzero algebraic submodule contains a K-type.

Write $g(t)=f(k_{2\pi t})$ for a smooth parity-$\epsilon$ function. Then $g(t+\frac{1}{2})=(-1{)}^{\epsilon }g(t)$, so $\widehat{g}(r)=0$ unless $r\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)$. By [F8], $\widehat{g^{(q)}}(r)=(2\pi ir{)}^q\widehat{g}(r)$ for every derivative order $q$, since the endpoint terms vanish by periodicity. Differentiating the finite Fourier sums in ${\sigma }_{N}g=(N+1{)}^{-1}{\sum }_{j=0}^N{S}_{j}g$ gives $({\sigma }_{N}g{)}^{(q)}={\sigma }_N(g^{(q)})$. Each ${\sigma }_{N}g$ is a finite sum of the allowed K-types, and [F7] applied to every $g^{(q)}$ shows ${\sigma }_{N}g\to g$ in every $C^q$ seminorm.

If $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$, neither coefficient in [F3] vanishes for an allowed weight $r$: either zero would force $\nu =-r-1$ or $\nu =r-1$, an integer congruent to $\epsilon +1$ modulo $2$. Repeated raising and lowering therefore connects every allowed weight to every other. By step 1.2 every nonzero algebraic submodule contains one weight, hence all of them, proving algebraic irreducibility.

Let $n\in {\mathcal{W}}_{\epsilon }$, $n\ge 1$, and set $\nu =n$. Put $T^{+}={⨁}_{j\ge 0}\mathbb{C}{f}_{n+1+2j}$, $T^{-}={⨁}_{j\ge 0}\mathbb{C}{f}_{-(n+1+2j)}$, and $F={⨁}_{j=0}^{n-1}\mathbb{C}{f}_{n-1-2j}$. The zeros $L_{E_{-}}{f}_{n+1}=0$ and $L_{E_{+}}{f}_{-n-1}=0$ make $T^{\pm }$ submodules; their internal arrows are nonzero, so each is simple by the weight-extraction argument of step 1.2. In the quotient by $T^{+}\oplus T^{-}$, the finite chain $F$ has nonzero internal arrows, and the class of $f_{n-1}$ is a highest-weight vector because its $E_{+}$ image lies in $T^{+}$. Its highest weight is $(n-1)\omega$, which is dominant integral since $n\ge 1$ and [F16]. The same interpolation as in step 1.2 extracts a weight from any nonzero submodule of this quotient; the internal arrows connect every weight of $F$, so $F$ is simple and [F12] identifies it with $L((n-1)\omega )=L_{n-1}$ in the notation of the Statement. By step 1.2, any submodule not contained in the two tails has a weight in $F$; its internal arrows reach all of $F$, and the arrows out of $f_{n-1}$ and $f_{1-n}$ have coefficient $n\mathrm{\ne }0$, so it then contains both tails. Consequently $T^{+}\oplus T^{-}$ is the unique maximal proper submodule and the finite quotient is the unique irreducible quotient. The filtration $0<T^{+}<T^{+}\oplus T^{-}<I_{\epsilon ,n}^K$ has three nonzero simple factors, so the composition length is $3$. Here $T^{+}\cong M_{n+1}^{-}$ is the lowest-weight string and $T^{-}\cong M_{-(n+1)}^{+}$ the highest-weight string.

If $\epsilon =1$ and $\nu =0$, then $L_{E_{-}}{f}_1=0$ and $L_{E_{+}}{f}_{-1}=0$. The positive odd chain ${⨁}_{j\ge 0}\mathbb{C}{f}_{1+2j}$ and negative odd chain ${⨁}_{j\ge 0}\mathbb{C}{f}_{-(1+2j)}$ are invariant; every internal arrow is nonzero, so each is simple by step 1.2. They have disjoint K-types and together contain every odd K-type, hence $I_{1,0}^K=M_1^{-}\oplus M_{-1}^{+}$.

Assume $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$ and let $W\mathrm{\ne }\{0\}$ be a closed $G$-invariant subspace of $H$. By [F4], $W\cap I_{\epsilon ,\nu }^K$ is a nonzero algebraic $(\mathfrak{g},K)$-submodule. Step 2.1 makes this intersection all of $I_{\epsilon ,\nu }^K$; its finite Fourier sums are dense in $H$ by [F2], so closedness gives $W=H$.

Assume $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$, and let $V\mathrm{\ne }\{0\}$ be a closed $G$-invariant subspace of $C_{\epsilon }^{\mathrm{\infty }}(K)$. Take $0\mathrm{\ne }f\in V$ and write $g(t)=f(k_{2\pi t})$. For an allowed $r$, form $P_{r}g(t)={\int }_0^1{e}^{-2\pi ir\varphi }g(t+\varphi )d\varphi$. Riemann sums lie in $V$ by $K$-invariance and converge in every $C^q$ seminorm: each ${\mathrm{\partial }}_t^q(e^{-2\pi ir\varphi }g(t+\varphi ))$ is uniformly continuous on the compact angle square by [F9], so the Riemann-sum error tends uniformly to zero in $t$. Periodicity and $s=t+\varphi$ give $P_{r}g(t)=\widehat{g}(r)e^{2\pi irt}$, hence $P_{r}g=\widehat{g}(r)f_r\in V$. Some allowed $\widehat{g}(r)$ is nonzero, since otherwise every Fejér mean of $g$ would vanish and [F7] would force $g=0$. Thus $V$ contains a K-type. For each real Lie algebra element, joint smoothness of the compact-picture cocycle implies that the derived difference quotients converge together with every angular derivative, hence in $C^{\mathrm{\infty }}$; closedness puts the derived vector in $V$, and complex-linear combinations give the raising and lowering operators in [F3]. Step 2.1's connected weight graph therefore puts every K-type in $V$, and step 1.3 plus closedness yields $V=C_{\epsilon }^{\mathrm{\infty }}(K)$.

For $\nu =-n$, the same finite chain $F$ is a submodule: its outward boundary arrows vanish, and its internal arrows are nonzero. Its highest-weight vector is $f_{n-1}$, killed by $E_{+}$, with highest weight $(n-1)\omega$, which is dominant integral since $n\ge 1$ and [F16]. By the same interpolation and internal-arrow argument as in step 2.2, it is simple, so [F12] gives $F\cong L((n-1)\omega )=L_{n-1}$. Modulo $F$, the positive and negative tails are separate submodules, because the crossing arrows land in $F$ and vanish in the quotient. Each has one-dimensional weight spaces and, by [F3], nonzero arrows in both directions between every adjacent pair of tail weights; only the inward boundary arrow vanishes in the quotient. The same interpolation as in step 1.2 extracts a weight from any nonzero submodule, and these internal raising and lowering arrows generate the whole tail, so both are simple. In either parameter, a one-sided string with fixed lowest (or highest) weight is unique up to rescaling its successive weight vectors: normalize each nonzero outward arrow to $1$, then $[E_{+},E_{-}]=W$ recursively fixes the inward arrows from the boundary condition. Thus the quotient factors are again $M_{n+1}^{-}$ and $M_{-(n+1)}^{+}$. Any irreducible submodule not contained in $F$ has a tail weight by step 1.2; repeated nonzero inward arrows first reach a tail boundary, whose inward arrow at $\nu =-n$ is nonzero into $F$. Its intersection with the simple submodule $F$ is then all of $F$, forcing the irreducible submodule to equal $F$. Hence $F$ is the unique irreducible submodule. If ${\tilde{T}}^{+}$ is the preimage of the positive quotient tail, the filtration $0<F<{\tilde{T}}^{+}<I_{\epsilon ,-n}^K$ has three nonzero simple factors, so the composition length is $3$.

By step 1.1, the Killing-dual basis gives $C=\frac{1}{8}{W}^2+\frac{1}{4}(E_{+}{E}_{-}+E_{-}{E}_{+})=\frac{1}{8}{W}^2-\frac{1}{4}W+\frac{1}{2}{E}_{+}{E}_{-}$ using [F11]. Formula [F3] yields $L_{E_{+}}{L}_{E_{-}}{f}_r=\frac{1}{4}({\nu }^2-(r-1{)}^2)f_r$, and substituting $L_W{f}_r=r{f}_r$ gives $C{f}_r=\frac{1}{8}({\nu }^2-1)f_r$ for every weight. Therefore $C$ acts by this scalar on $I_{\epsilon ,\nu }^K$ and on its subquotient $L_{n-1}$ at $\nu =n$.

Use the rank-one Cartan, root, coroot, positive-system and Weyl data from step 1.1. In PBW order $E_{-}<W<E_{+}$, $E_{+}{E}_{-}=E_{-}{E}_{+}+W$, so the Harish-Chandra projection of the Casimir is $\mathrm{pr}(C)=W^2\mathrm{/}8+W\mathrm{/}4$; hence the shifted polynomial is ${\mathrm{HC}}_{\rho }(C)(t)=(t^2-1)\mathrm{/}8$. This generates $S(\mathfrak{h}{)}^W=\mathbb{C}[t^2]$, so [F11] implies $Z(U(\mathfrak{g}))=\mathbb{C}[C]$. Since $C$ acts by ${\lambda }_{\nu }=({\nu }^2-1)\mathrm{/}8$, every $p(C)$ acts by $p({\lambda }_{\nu })$ and defines the central character of [F13]; two such characters are equal exactly when ${\nu }^2=({\nu }^{\mathrm{\prime }}{)}^2$, that is, when ${\nu }^{\mathrm{\prime }}=\pm \nu$.